Question

From each end of a solid metal cylinder, metal was scooped out in hemispherical from a same diameter. The height of the cylinder is $10cm$ and its base of its radius $4.2cm$. The rest of the cylinder is melted and converted into a cylindrical wire of $1.4cm$ thickness. Find the length of the wire.

Answer 1

Height of the solid metal cylinder $h=10$cm

Radius of the solid metal cylinder $r=4.2$cm

$\therefore$ Radius of each hemisphere$=$ radius of the solid metal cylinder,$r=4.2$cm

Now, Volume of the rest of the cylinder$=$Volume of cylinder$-2\times$Volume of each hemisphere.

$=\pi r^{2}h-2\times \frac{2}{3}\pi r^3$

$=\pi r^2(h-\frac{4r}{3})$

$=\pi \times {4.2}^2(10-\frac{4\times 4.2}{3})$

$=\pi \times {4.2}^2\times 4.4{cm}^3$

Thickness of the cylindrical wire$=1.4$cm

$\therefore$Radius of the cylindrical wire$=R=\frac{1.4}{2}$cm$=0.7$cm

Let the length of the wire be $H$cm

It is given that the rest of the cylinder is melted and converted in to a cylindriacal wire.

$\therefore$ Volume of the cylindrical wire$=$Volume of the rest of the cylinder

$\Rightarrow \pi \times 0.7\times 0.7\times H=\pi \times {\left(4.2\right)}^2\times 4.4$

$\Rightarrow H=\frac{4.2\times 4.2\times 4.4}{0.7\times 0.7}=158.4$cm

Hence, the length of the wire is $158.4$cm